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Trial and error https://www.calcudoku.org/forum/viewtopic.php?f=2&t=65 |
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Author: | pharosian [ Sun Oct 02, 2011 8:18 pm ] |
Post subject: | Re: Trial and error |
pnm wrote: Given that the 16x could be either 4x1x4 or 2x4x2 and that 1- could be either 5-3-1, 5-2-2, or 4-2-1 (each in various combinations within the cage), I fail to see how "you can figure out quite easily that there should be a 3 in C5." To me, it looks like a lot of trial and error just to get to that point. Also, earlier in this thread a couple of people mention "parity checks." What is that, and how do you use it to solve calcudoku? |
Author: | pnm [ Sun Oct 02, 2011 9:32 pm ] |
Post subject: | Re: Trial and error |
pharosian wrote: Given that the 16x could be either 4x1x4 or 2x4x2 and that 1- could be either 5-3-1, 5-2-2, or 4-2-1 (each in various combinations within the cage), I fail to see how "you can figure out quite easily that there should be a 3 in C5." To me, it looks like a lot of trial and error just to get to that point. Well, yes, you 're right, you would have to try those combinations. pharosian wrote: Also, earlier in this thread a couple of people mention "parity checks." What is that, and how do you use it to solve calcudoku? Check out clm's thread on "the parity" Patrick |
Author: | clm [ Mon Oct 03, 2011 12:34 am ] |
Post subject: | Re: Trial and error |
pnm wrote: pharosian wrote: Given that the 16x could be either 4x1x4 or 2x4x2 and that 1- could be either 5-3-1, 5-2-2, or 4-2-1 (each in various combinations within the cage), I fail to see how "you can figure out quite easily that there should be a 3 in C5." To me, it looks like a lot of trial and error just to get to that point. Well, yes, you 're right, you would have to try those combinations. pharosian wrote: Also, earlier in this thread a couple of people mention "parity checks." What is that, and how do you use it to solve calcudoku? Check out clm's thread on "the parity" Patrick A quick way to see this: Although the combination 3-1-1 is also possible for the cage 1-, it is rapidly supressed since 16x should be 2-4-2 and now the total known sum would be 5 + 11 + 8 = 24 so a 6 (30 - 24 to complete 2 rows) would go to E4. Then, if a 3 is inside the cage -1, the other two numbers must be 1 and 5 necessarily and now 9 + 11 = 20 and the sum of 16x and the cell E4 equals 10, but 2-4-2 and a 2 in E4 is not possible and 4-4-1 and a 1 in E4 go against the hypothesis (because in this case two 1's have already been considered present in rows 4 and 5). So the 3 must go to C5. (and, inmediately, B4 = 3, C4 = 5). Also we see that 2-4-2 is not valid in 16x since 5-1 in A5-B5 produce a 3 in A4,... , concluding that 16x = 4-1-4 and then 5-2-2 for the cage 1-, etc.). But perhaps the fastest way is: To consider only 3-1-1 (as before), supressing this combination, and since 4-2-1 is not possible (two 4's in case of 4-1-4 or two 2's in case of 2-4-2), the other possibilities 5-3-1 or 5-2-2 have a sum of 9, so the sum of 16x and E4 equals 10, then 2-4-2 is not valid (E4 = 2) so 16x = 4-1-4 and E4 = 1, and now two 1's in rows 4 and 5 drive to 1- = 5-2-2 (and not 5-3-1). Rows 4 and 5 are totally defined. For the question of the parity, and using this same diagram: The sum of the whole puzzle is odd (5 x 15 = 75). Every two cages odd produce a sum of all numbers inside them even (for instance the cages 1- and 11+); 2- we know is an even cage, and since 3- is odd, finally the addition of the cages 2: and 16x must be even, but now we know that 16x is odd (4-1-4) so 2: must be odd too and then 1-2 is the only valid combination (not 2-4); we can write the pair 1-2 as the final candidates in B1-C1 inmediately after solving the cage 16x. Thanks, Patrick, for the reference to my original post, in that moment I did not provide a diagram showing a good example of the use of this property, but anyway the target is understanding the nuclear idea. |
Author: | pharosian [ Tue Oct 04, 2011 3:56 pm ] |
Post subject: | Re: Trial and error |
Thanks, clm! I have since read your article on parity. I'm still trying to get used to thinking about cages in these terms, but I see how it could be helpful. Of more immediate assistance, however, is your comment about keeping the sum of the rows (or columns) in mind. I use a sum on each row and column when I'm using Excel to solve book puzzles, but it's always been more of a "sanity check" to ensure that my solution doesn't have dupes in a row or column because my Excel spreadsheet doesn't have Patrick's handy "continuous checking" property to flag errors. I see now, though, that if I keep the sum in mind as you describe in your first sentence, I could rule out possibilities more quickly. |
Author: | jomapil [ Tue Oct 11, 2011 7:23 pm ] |
Post subject: | Re: Trial and error |
Patrick, the 5x5 difficult of today (11OCT11) I only "analysed" the row 5 : 34,34,125,125,125 After that, it is impossible ( for me ) to continue the solution without the Trial and Error. Did I fail anything or are there any of those little puzzles only solved with Trial and Error? |
Author: | pnm [ Tue Oct 11, 2011 8:08 pm ] |
Post subject: | Re: Trial and error |
jomapil wrote: After that, it is impossible ( for me ) to continue the solution without the Trial and Error. Did I fail anything or are there any of those little puzzles only solved with Trial and Error? The way I'd continue would be along these lines: - the 1- in e1,e2,e3 can only be 4,1,2 or 5,1,3 - so the number in e5 has to be a 2 or a 5 - so the number in e4 has to be a 3 or a 4 - so the 8+ cage must have a 3 somewhere then: - the 2- cage in a3,a4,b4 can only be 4,1,1 or 5,1,2 - the 11+ cage can only be 2,4,5 or 5,1,5 (3,5,3 not possible because of the 3 in the 8+ cage) - but 5,1,5 also impossible, because then the 2- cage should be 4,1,1 - so the 11+ cage must have 2,4,5 then: - the 2- cage cannot be 4,1,1 because then the left column would have 4,1,3, leaving 2,5 for the 2- cage at the top left so: - the 8+ cage cannot have 2,3 in d3 and d4 - so a 1 has to go into c5 etc. etc. I'll leave it to others to finish the story Patrick |
Author: | sneaklyfox [ Tue Oct 11, 2011 10:50 pm ] |
Post subject: | Re: Trial and error |
I did it more like this: - after 3,4 and 1,2,5 in row 5... - a3,a4,b4 has to be 1,1,4 or 1,2,5 - it cannot be 1,1,4 because then a5,b5 would be 3,4 which only leaves 2 and 5 to go into the 2- cage at a1,a2 - so a3,a4,b5 must be 1,2,5 - then sum row 3 and 4 = 1+2+5+11+8+e3 = 30 so e3=3 - e1,e2,e3 can only be 1,3,5 with e3=3 - then e5=2, e4=4 - then d3, d4 must be 1,3 with d3=1, d4=3 - then d5=5, c5=1 - then d1,d2 must be 4,2 so c1=3 - since solutions are unique, a1,a2 cannot be 2,4 so there must be a2=3 - then a5=4, b5=3 - there can only be 5,2,2 or 3,1,1 or 4,1,2 or 5,1,3 in 1- L-shaped cage so b1,b2,c2 must be 4,1,2 or 5,1,3 (must have a 1) - so a1=5, e1=1, e2=5, a3=2, a4=1, b4=5 - then b1,b2,c2 must be 4,1,2 - since solutions are unique, b2=1 - then b1=2, b3=4, c2=4, c3=5, c4=2 DONE Lots of other ways to arrive at solution... |
Author: | pnm [ Tue Oct 11, 2011 10:56 pm ] |
Post subject: | Re: Trial and error |
sneaklyfox wrote: I did it more like this: - after 3,4 and 1,2,5 in row 5... ... - then b1=2, b3=4, c2=4, c3=5, c4=2 DONE This shows why you're always faster than me |
Author: | jomapil [ Tue Oct 11, 2011 11:11 pm ] |
Post subject: | Re: Trial and error |
I never performed an analysis in that way. It looks like it is appropriate for little puzzles. With 3 weeks as a calcudolic I conclude the resolution of a 5x5 puzzle is a little different that a 9x9 or a 12x12, though the principles are the same. Thank you Patrick and sneaklyfox for the lesson. |
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